MathIsimple

Solid Geometry – Problem 15: Find the dihedral angle along edge between planes and

Question

In a regular square pyramid PABCDP-ABCD, the base ABCDABCD is a square of side 22. The apex PP is directly above the center OO of the base, and PO=2PO=2. Find the dihedral angle along edge ABAB between planes PABPAB and ABCDABCD.

Step-by-step solution

Set up coordinates with the base in z=0z=0 and center at the origin: A(1,1,0), B(1,1,0), C(1,1,0), D(1,1,0), P(0,0,2).A(-1,-1,0),\ B(1,-1,0),\ C(1,1,0),\ D(-1,1,0),\ P(0,0,2). Plane ABCDABCD is z=0z=0, so a normal vector is n1=(0,0,1)\vec n_1=(0,0,1). For plane PABPAB, use two direction vectors in the plane: AB=(2,0,0),AP=(1,1,2).\overrightarrow{AB}=(2,0,0),\qquad \overrightarrow{AP}=(1,1,2). A normal vector is n2=AB×AP=(0,4,2)(0,2,1).\vec n_2=\overrightarrow{AB}\times\overrightarrow{AP}=(0,-4,2)\sim(0,-2,1). Let θ\theta be the dihedral angle (the angle between the two planes). Then \cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}= rac{1}{\sqrt5}.

Final answer

The dihedral angle θ\theta satisfies cosθ=15\cos\theta=\dfrac{1}{\sqrt5} (so θ=arccos15\theta=\arccos\dfrac{1}{\sqrt5}).

Marking scheme

1. Checkpoints (max 7 pts total)

  • Coordinate setup (2 pts): Place the square base of side 22 in z=0z=0 with correct center and height 22.
  • Normal vectors (3 pts): Find n1\vec n_1 for the base plane and n2=AB×AP\vec n_2=\overrightarrow{AB}\times\overrightarrow{AP} for plane PABPAB.
  • Angle computation (2 pts): Compute cosθ=n1n2n1n2=15\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=\frac{1}{\sqrt5}.

2. Zero-credit items

  • Using a 2D triangle angle in a face as the dihedral angle without justification.
  • Giving cosθ\cos\theta without any normal-vector work.

3. Deductions

  • Cross-product mistake (-1): incorrect AB×AP\overrightarrow{AB}\times\overrightarrow{AP}.
  • Normalization/arithmetic error (-1): incorrect computation of n2|\vec n_2|.
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